Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

bsort(nil) → nil
bsort(.(x, y)) → last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) → nil
bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) → 0
last(.(x, nil)) → x
last(.(x, .(y, z))) → last(.(y, z))
butlast(nil) → nil
butlast(.(x, nil)) → nil
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

Q restricted rewrite system:
The TRS R consists of the following rules:

bsort(nil) → nil
bsort(.(x, y)) → last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) → nil
bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) → 0
last(.(x, nil)) → x
last(.(x, .(y, z))) → last(.(y, z))
butlast(nil) → nil
butlast(.(x, nil)) → nil
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

bsort(nil) → nil
bsort(.(x, y)) → last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) → nil
bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) → 0
last(.(x, nil)) → x
last(.(x, .(y, z))) → last(.(y, z))
butlast(nil) → nil
butlast(.(x, nil)) → nil
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

BUBBLE(.(x, .(y, z))) → BUBBLE(.(x, z))
BSORT(.(x, y)) → BSORT(butlast(bubble(.(x, y))))
BSORT(.(x, y)) → LAST(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
LAST(.(x, .(y, z))) → LAST(.(y, z))
BUTLAST(.(x, .(y, z))) → BUTLAST(.(y, z))
BSORT(.(x, y)) → BUTLAST(bubble(.(x, y)))
BUBBLE(.(x, .(y, z))) → BUBBLE(.(y, z))
BSORT(.(x, y)) → BUBBLE(.(x, y))

The TRS R consists of the following rules:

bsort(nil) → nil
bsort(.(x, y)) → last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) → nil
bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) → 0
last(.(x, nil)) → x
last(.(x, .(y, z))) → last(.(y, z))
butlast(nil) → nil
butlast(.(x, nil)) → nil
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

Q DP problem:
The TRS P consists of the following rules:

BUBBLE(.(x, .(y, z))) → BUBBLE(.(x, z))
BSORT(.(x, y)) → BSORT(butlast(bubble(.(x, y))))
BSORT(.(x, y)) → LAST(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
LAST(.(x, .(y, z))) → LAST(.(y, z))
BUTLAST(.(x, .(y, z))) → BUTLAST(.(y, z))
BSORT(.(x, y)) → BUTLAST(bubble(.(x, y)))
BUBBLE(.(x, .(y, z))) → BUBBLE(.(y, z))
BSORT(.(x, y)) → BUBBLE(.(x, y))

The TRS R consists of the following rules:

bsort(nil) → nil
bsort(.(x, y)) → last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) → nil
bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) → 0
last(.(x, nil)) → x
last(.(x, .(y, z))) → last(.(y, z))
butlast(nil) → nil
butlast(.(x, nil)) → nil
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

BSORT(.(x, y)) → BSORT(butlast(bubble(.(x, y))))
BUBBLE(.(x, .(y, z))) → BUBBLE(.(x, z))
BSORT(.(x, y)) → LAST(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
LAST(.(x, .(y, z))) → LAST(.(y, z))
BSORT(.(x, y)) → BUTLAST(bubble(.(x, y)))
BUTLAST(.(x, .(y, z))) → BUTLAST(.(y, z))
BUBBLE(.(x, .(y, z))) → BUBBLE(.(y, z))
BSORT(.(x, y)) → BUBBLE(.(x, y))

The TRS R consists of the following rules:

bsort(nil) → nil
bsort(.(x, y)) → last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) → nil
bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) → 0
last(.(x, nil)) → x
last(.(x, .(y, z))) → last(.(y, z))
butlast(nil) → nil
butlast(.(x, nil)) → nil
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 3 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

BUTLAST(.(x, .(y, z))) → BUTLAST(.(y, z))

The TRS R consists of the following rules:

bsort(nil) → nil
bsort(.(x, y)) → last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) → nil
bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) → 0
last(.(x, nil)) → x
last(.(x, .(y, z))) → last(.(y, z))
butlast(nil) → nil
butlast(.(x, nil)) → nil
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


BUTLAST(.(x, .(y, z))) → BUTLAST(.(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
BUTLAST(x1)  =  BUTLAST(x1)
.(x1, x2)  =  .(x2)

Lexicographic Path Order [19].
Precedence:
.1 > BUTLAST1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

bsort(nil) → nil
bsort(.(x, y)) → last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) → nil
bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) → 0
last(.(x, nil)) → x
last(.(x, .(y, z))) → last(.(y, z))
butlast(nil) → nil
butlast(.(x, nil)) → nil
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

LAST(.(x, .(y, z))) → LAST(.(y, z))

The TRS R consists of the following rules:

bsort(nil) → nil
bsort(.(x, y)) → last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) → nil
bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) → 0
last(.(x, nil)) → x
last(.(x, .(y, z))) → last(.(y, z))
butlast(nil) → nil
butlast(.(x, nil)) → nil
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


LAST(.(x, .(y, z))) → LAST(.(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LAST(x1)  =  LAST(x1)
.(x1, x2)  =  .(x2)

Lexicographic Path Order [19].
Precedence:
.1 > LAST1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

bsort(nil) → nil
bsort(.(x, y)) → last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) → nil
bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) → 0
last(.(x, nil)) → x
last(.(x, .(y, z))) → last(.(y, z))
butlast(nil) → nil
butlast(.(x, nil)) → nil
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

BUBBLE(.(x, .(y, z))) → BUBBLE(.(x, z))
BUBBLE(.(x, .(y, z))) → BUBBLE(.(y, z))

The TRS R consists of the following rules:

bsort(nil) → nil
bsort(.(x, y)) → last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) → nil
bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) → 0
last(.(x, nil)) → x
last(.(x, .(y, z))) → last(.(y, z))
butlast(nil) → nil
butlast(.(x, nil)) → nil
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


BUBBLE(.(x, .(y, z))) → BUBBLE(.(x, z))
BUBBLE(.(x, .(y, z))) → BUBBLE(.(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
BUBBLE(x1)  =  BUBBLE(x1)
.(x1, x2)  =  .(x2)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

bsort(nil) → nil
bsort(.(x, y)) → last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) → nil
bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) → 0
last(.(x, nil)) → x
last(.(x, .(y, z))) → last(.(y, z))
butlast(nil) → nil
butlast(.(x, nil)) → nil
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

BSORT(.(x, y)) → BSORT(butlast(bubble(.(x, y))))

The TRS R consists of the following rules:

bsort(nil) → nil
bsort(.(x, y)) → last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) → nil
bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) → 0
last(.(x, nil)) → x
last(.(x, .(y, z))) → last(.(y, z))
butlast(nil) → nil
butlast(.(x, nil)) → nil
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.